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\title{CS 5846 Decision Theory I: Assignment 3} 
\author{Yifan Tong (UG), yt347} 
\date{November 11, 2010}

\begin{document} 
\maketitle 
\newpage
\section{Problem 1} % (fold)
\label{sec:problem_1}
\subsection{Part a}
Given $u(x) = x - 0.005x^2$,
\[E_{p_1}(u(x))=\frac{1}{4}u(30) + \frac{1}{4}u(50)+\frac{1}{4}u(70) = 38.5\]
\[E_{p_2}(u(x))=u(30)=25.5\]
Therefore, $p_1\succ p_2$
\subsection{Part b}
Given $u(x) = x - 0.01x^2$,
\[E_{p_1}(u(x))=\frac{1}{4}u(30) + \frac{1}{4}u(50)+\frac{1}{4}u(70) = 22\]
\[E_{p_2}(u(x))=u(30)=21\]
Therefore, $p_1\succ p_2$

\subsection{Part c}
\begin{eqnarray*}
E_{p_1}(u(x))&=&E_{p_1}(u(x))\\
\frac{1}{4}(30-900b) + \frac{1}{4}(50-2500b) + \frac{1}{2}(70-4900b) &=& 30-900b\\
b &=& \frac{1}{96}
\end{eqnarray*}
Decision maker is indifferent if $b=\frac{1}{96}$. Decision maker prefers $p_1$ for $b<\frac{1}{96}$; $p_2$, otherwise.
\begin{center}
\includegraphics[scale=0.8]{graph3.png}
\end{center}


% section problem_1 (end)

\section{Problem 2} % (fold)
\label{sec:problem_2}
\textbf{Independence Theorem}:\\
For some $p\succ q$; $p, q, r\in P$
\[\alpha p + (1-\alpha)r \succ \alpha q +(1-\alpha)r\]
Given that preferences $\succ$ have an expected utility function, $u$, $p\succ q$ iff
\[\sum_{z\in Z}u(z)p(z) > \sum_{z\in Z}u(z)q(z)\]
Assume, for contradiction, $\succ$ does not satisfy independence theorem:\\
\[\sum_{z\in Z}u(z)[\alpha p(z)+(1-\alpha)r(z)] \leq \sum_{z\in Z}u(z)[\alpha q(z)+(1-\alpha)r(z)]\]
\[\sum_{z\in Z}u(z)\alpha p(z)+\sum_{z\in Z}u(z)(1-\alpha)r(z) \leq \sum_{z\in Z}u(z)\alpha q(z)+\sum_{z\in Z}u(z)(1-\alpha)r(z)\]
\[\sum_{z\in Z}u(z)\alpha p(z) \leq \sum_{z\in Z}u(z)\alpha q(z)\]
\[\sum_{z\in Z}u(z)p(z) \leq \sum_{z\in Z}u(z)q(z)\]
We have a contradiction, since, by definition, $\sum_{z\in Z}u(z)p(z) > \sum_{z\in Z}u(z)q(z)$.  Therefore, $\succ$ does satisfy independence theorem.
% section problem_2 (end)

\section{Problem 3} % (fold)
\label{sec:problem_3}
Given $u$ represents $\succ$:
\[p\succ q \leftrightarrow \sum_{z\in Z}u(z)p(z) > \sum_{z\in Z}u(z)q(z)\]
Given $v(\cot)=au(\cot)+b$, need to show two things
\subsection{$\rightarrow$}
\textbf{Prove}: $p\succ q \rightarrow \sum_{z\in Z}v(z)p(z) > \sum_{z\in Z}v(z)q(z)$\\\\
Assume, for contradiction, $\sum_{z\in Z}v(z)p(z) \leq \sum_{z\in Z}v(z)q(z)$ given $p\succ q$
\begin{eqnarray*}
\sum_{z\in Z}v(z)p(z) &\leq& \sum_{z\in Z}v(z)q(z)\\
\sum_{z\in Z}[au(z)+b]p(z) &\leq& \sum_{z\in Z}[au(z)+b]q(z)\\
a\sum_{z\in Z}u(z)p(z)+b\sum_{z\in Z}p(z) &\leq& a\sum_{z\in Z}u(z)q(z)+b\sum_{z\in Z}q(z)\\
a\sum_{z\in Z}u(z)p(z)+b &\leq& a\sum_{z\in Z}u(z)q(z)+b\\
\sum_{z\in Z}u(z)p(z) &\leq& \sum_{z\in Z}u(z)q(z)
\end{eqnarray*}
We have reached a contradiction since $p\succ q \leftrightarrow \sum_{z\in Z}u(z)p(z) > \sum_{z\in Z}u(z)q(z)$.\\\\
Therefore, $p\succ q \rightarrow \sum_{z\in Z}v(z)p(z) > \sum_{z\in Z}v(z)q(z)$
\subsection{$\leftarrow$}
\textbf{Prove}: $p\succ q \leftarrow \sum_{z\in Z}v(z)p(z) > \sum_{z\in Z}v(z)q(z)$\\\\
Assume, for contradiction $\sum_{z\in Z}v(z)p(z) > \sum_{z\in Z}v(z)q(z) \rightarrow  p\not\succ q$
\begin{eqnarray*}
\sum_{z\in Z}v(z)p(z) &>& \sum_{z\in Z}v(z)q(z)\\
\sum_{z\in Z}[au(z)+b]p(z) &>& \sum_{z\in Z}[au(z)+b]q(z)\\
a\sum_{z\in Z}u(z)p(z)+b\sum_{z\in Z}p(z) &>& a\sum_{z\in Z}u(z)q(z)+b\sum_{z\in Z}q(z)\\
a\sum_{z\in Z}u(z)p(z)+b &>& a\sum_{z\in Z}u(z)q(z)+b\\
\sum_{z\in Z}u(z)p(z) &>& \sum_{z\in Z}u(z)q(z)\\
p &\succ& q
\end{eqnarray*}
Again, we have reached a contradiction; therefore, $p\succ q \leftarrow \sum_{z\in Z}v(z)p(z) > \sum_{z\in Z}v(z)q(z)$\\\\
We can safely conclude that $v$ also represents $\succ$.
% section problem_3(end)


\section{Problem 4} % (fold)
\label{sec:problem_4}
(Grad)
% section problem_4(end)

\section{Problem 5} % (fold)
\label{sec:problem_5}
\subsection{Part a}
In event of a loss:
\[\mbox{total wealth }=w-a-rx+x\]
\subsection{Part b}
No loss:
\[\mbox{total wealth }=w-rx\]
\subsection{Part c}
\begin{eqnarray*}
w-a-rx+x &=& w - rx\\
x &=& a
\end{eqnarray*}
\subsection{Part d}
\[E = p*u(w-a-rx+x)+(1-p)*u(w-rx)\]
To find optimal, take the derivative
\[\frac{dE}{dx}=p(1-r)u'(w-a-rx+x) + (1-p)(-r)u'(w-rx)\]
To ensure this is in fact maximum, take the second derivative,
\[\frac{d^2E}{dx^2}=p(1-r)^2u''(w-a-rx+x) + (1-p)r^2u''(w-rx)\]
Since $p>0$, $1-p>0$, $(1-r)^2> 0$, $r^2>0$, and $u''(\cdot)<0$,
\[\frac{d^2E}{dx^2}<0\]
Therefore, solving $\frac{dE}{dx}=0$ yields a maximum; solve with $x=a$,
\begin{eqnarray*}
p(1-r)u'(w-a-rx+x) + (1-p)(-r)u'(w-rx)&=&0\\
p(1-r)u'(w-a-ra+a) + (1-p)(-r)u'(w-ra)&=&0\\
p(1-r)u'(w-ra) + (1-p)(-r)u'(w-ra)&=&0\\
(p)u'(w-ra)- (p)(r)u'(w-ra) -(r)u'(w-ra)+(p)(r)u'(w-ra)&=&0\\
(p)u'(w-ra) -(r)u'(w-ra)&=&0\\
(p)u'(w-ra) &=&(r)u'(w-ra)\\
r &=& p
\end{eqnarray*}
Therefore, she will demand $r=p$.

% section problem_5 (end)

\section{Problem 6} % (fold)
\label{sec:problem_6}
\subsection{Part a}
\[\tilde y \sim N(m+xr, x^2\sigma^2)\]
\subsection{Part b}
\[E(u(y)) = -e^{-\lambda (m+xr - \frac{\lambda}{2}x^2\sigma^2)}\]
\subsection{Part c}

\subsubsection{$\frac{\partial E}{\partial x}$}
\[\frac{\partial E}{\partial x} = (-\lambda r + \lambda^2x\sigma^2 )E\]
To ensure this is a maximum, take the second derivative,
\[\frac{\partial^2 E}{\partial x^2} = \lambda^2\sigma^2E + (-\lambda r + \lambda^2x\sigma^2 )^2E\]
Observe that
\begin{eqnarray*}
\lambda^2 &>&0\\
\sigma^2&>&0\\
(-\lambda r + \lambda^2x\sigma^2 )^2&>&0\\
E&<&0
\end{eqnarray*}
Therefore, $\frac{\partial^2 E}{\partial x^2} < 0$; and therefore, $\frac{\partial E}{\partial x} = 0$ produces a maximum.\\\\
Solve,
\begin{eqnarray*}
\frac{\partial E}{\partial x} &=& 0\\
(-\lambda r + \lambda^2x\sigma^2 ) &=& 0\\
x &=&\frac{r}{\lambda\sigma^2}
\end{eqnarray*}

\subsubsection{$\frac{\partial E}{\partial m}$}
Given $E = -e^{-\lambda (m+xr - \frac{\lambda}{2}x^2\sigma^2)}$
\[\frac{\partial E}{\partial m} = -\lambda E\]
\[\frac{\partial^2 E}{\partial m^2} = \lambda^2 E\]
Since $E<0$, $\lambda\not = 0$, $\frac{\partial E}{\partial m} \not = 0$; there are no local maxima of $E$ relative to $m$.  However, since $\frac{\partial^2 E}{\partial m^2} = \lambda^2 E < 0$, $\frac{\partial E}{\partial m}$ is optimal if we make $m$ as large as possible.\\
Since $x =\frac{r}{\lambda\sigma^2}$ is optimal, $m = w_0 - x$ is also optimal.

\subsubsection{Optimal $E$}
Given,
\[x =\frac{r}{\lambda\sigma^2}\]
\[m = w_0 - \frac{r}{\lambda\sigma^2}\]
The optimal portfolio is defined as follows,
\begin{eqnarray*}
E_{opt} &=& E(r, \sigma^2)\\
&=& -e^{-\lambda(w_0 - \frac{r}{\lambda\sigma^2} + \frac{r}{\lambda\sigma^2}r - \frac{\lambda}{2}(\frac{r}{\lambda\sigma^2})^2\sigma^2)}\\
&=& -e^{-\lambda[w_0 - (1-r)\frac{r}{\lambda\sigma^2} - \frac{\lambda}{2}(\frac{r}{\lambda\sigma^2})^2\sigma^2]}
\end{eqnarray*}
% section problem_6 (end)

\section{Problem 7} % (fold)
\label{sec:problem_7}

% section problem_7 (end)
\subsection{Part a}
Given $u(x)=\gamma^{-1}x^{-\gamma}$
\[\frac{1}{2}[u(85)+u(115)]=u(p^*)\]
\[p^* = (\frac{2}{85^{-\gamma}+115^{-\gamma}})^{\frac{1}{\gamma}}\]
For $\gamma=1$,
\[p^*=\frac{391}{4}=97.75\]
For $\gamma=0.5$,
\[p^*\approx 98.3061\]
For $\gamma=-1$,
\[p^*=100\]
\subsection{Part b}
Plot of $p(\gamma)$, for $\gamma=-100\ldots 100$\\
Notice that $\gamma\rightarrow\infty$, $p^*\rightarrow 85$; $\gamma\rightarrow-\infty$, $p^*\rightarrow 115$.  This is not surprising because for $\gamma>0$, $85^{-\gamma}$ is the more dominant term, whereas for $\gamma<0$, $115^{-\gamma}$.
\begin{center}
\includegraphics[scale=0.8]{graph1.png}
\end{center}
For arbitrary $\gamma\leq1$, $p^*$ diminishes rather quickly until 112, and further diminishes towards 115 at a very slow speed.  
\begin{center}
\includegraphics[scale=0.8]{graph2.png}
\end{center}

\end{document}